I cannot seem to understand why the function $F$ defined in Theorem 7.1 of the paper “Permutation rotation-symmetric Sboxes, liftings and affine equivalence” is described as “a bijection on $\mathbb{F}_2^n$”.

The input contains $n$ bits, yet the given definition seems to imply that the output contains $k=n-2$ bits: $$F(x_1, x_2, \ldots, x_n) = (f(x_1, \ldots, x_k), f(x_2, \ldots, x_{k+1}), \ldots, f(x_k, x_1, \ldots, x_{k-1})).$$

There is absolutely no way that such function can be bijective, so I must be missing some essential detail.

For example, can anyone demonstrate how to compute the value of, say, $F(00001)$?

  • $\begingroup$ Note that being ’bijective’ doesn’t imply that there is an easy or even a known way to calculate both ways. $\endgroup$
    – Aganju
    Mar 11, 2022 at 20:55

1 Answer 1


It is simply a typo in the paper I believe. It should say: $$F(x_1, x_2, \ldots, x_n) = (f(x_1, \ldots, x_k), f(x_2, \ldots, x_{k+1}), \ldots, f(x_n, x_1, \ldots, x_{k-1})).$$

(Note the $x_n$ instead of $x_k$ in the final evaluation of $f$). This is what was written on page 1 of the paper, and has $n$-bit output.


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